2. The acceleration of a simple harmonic oscillator is zero whenever the oscillating object is at the equilibrium position.

Transcription

1 CHAPER : Vibratins and Waes Answers t Questins. he blades in an electric shaer ibrate, apprximately in SHM. he speakers in a stere system ibrate, but usually in a ery cmplicated way since many ntes are being sunded at the same time. A pian string ibrates when struck, in apprximately SHM. he pistns in a car engine scillate, in apprximately SHM. he free end f a diing bard scillates after a dier jumps, in apprximately SHM.. he acceleratin f a simple harmnic scillatr is zer wheneer the scillating bject is at the equilibrium psitin. 3. he mtin f the pistn can be apprximated as simple harmnic. First f all, the pistn will hae a cnstant perid while the engine is running at a cnstant speed. he speed f the pistn will be zer at the extremes f its mtin the tp and bttm f the strke which is the same as in simple harmnic mtin. here is a large frce exerted n the pistn at ne extreme f its mtin the cmbustin f the fuel mixture and simple harmnic mtin has the largest frce at the extremes f the mtin. Als, as the crankshaft mes in a circle, its cmpnent f mtin in ne dimensin is transferred t the pistn. It is similar t Fig Since the real spring has mass, the mass that is ming is greater than the mass at the end f the k spring. Since f, a larger mass means a smaller frequency. hus the true frequency will m be smaller than the massless spring apprximatin. And since the true frequency is smaller, the true perid will be larger than the massless spring apprximatin. Abut /3 the mass f the spring cntributes t the ttal mass alue. 5. he imum speed is gien by A k m. Varius cmbinatins f changing A, k, and/r m can result in a dubling f the imum speed. Fr example, if k and m are kept cnstant, then dubling the amplitude will duble the imum speed. Or, if A and k are kept cnstant, then reducing the mass t ne-furth its riginal alue will duble the imum speed. Nte that changing either k r m will als change the frequency f the scillatr, since k f. m 6. he scale reading will scillate with damped scillatins abut an equilibrium reading f 5.0 kg, with an initial amplitude f 5.0 kg (s the range f readings is initially frm 0.0 kg and 0.0 kg). Due t frictin in the spring and scale mechanism, the scillatin amplitude will decrease er time, eentually cming t rest at the 5.0 kg mark. 7. he perid f a pendulum clck is inersely prprtinal t the square rt f g, by Equatin -a, g. When taken t high altitude, the alue f g will decrease (by a small amunt), which means the perid will increase. If the perid is t lng, the clck is running slw and s will lse time. 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 69

2 Chapter Vibratins and Waes 8. he tire swing apprximates a simple pendulum. With a stpwatch, yu can measure the perid f g the tire swing, and then sle Equatin -a fr the length, make the water slsh, yu must shake the water (and the pan) at the natural frequency fr water waes in the pan. he water then is in resnance, r in a standing wae pattern, and the amplitude f scillatin gets large. hat natural frequency is determined by the size f the pan smaller pans will slsh at higher frequencies, crrespnding t shrter waelengths fr the standing waes. he perid f the shaking must be the same as the time it takes a water wae t make a rund trip in the pan. 0. Sme examples f resnance: Pushing a child n a playgrund swing yu always push at the frequency f the swing. Seeing a stp sign scillating back and frth n a windy day. When singing in the shwer, certain ntes will sund much luder than thers. Utility lines alng the radside can hae a large amplitude due t the wind. Rubbing yur finger n a wineglass and making it sing. Blwing acrss the tp f a bttle. A rattle in a car (see Questin ).. A rattle in a car is ery ften a resnance phenmenn. he car itself ibrates in many pieces, because there are many peridic mtins ccurring in the car wheels rtating, pistns ming up and dwn, ales pening and clsing, transmissin gears spinning, drieshaft spinning, etc. here are als ibratins caused by irregularities in the rad surface as the car is drien, such as hitting a hle in the rad. If there is a lse part, and its natural frequency is clse t ne f the frequencies already ccurring in the car s nrmal peratin, then that part will hae a larger than usual amplitude f scillatin, and it will rattle. his is why sme rattles nly ccur at certain speeds when driing.. he frequency f a simple peridic wae is equal t the frequency f its surce. he wae is created by the surce ming the wae medium that is in cntact with the surce. If yu hae ne end f a taut string in yur hand, and yu me yur hand with a frequency f Hz, then the end f the string in yur hand will be ming at Hz, because it is in cntact with yur hand. hen thse parts f the medium that yu are ming exert frces n adjacent parts f the medium and cause them t scillate. Since thse tw prtins f the medium stay in cntact with each ther, they als must be ming with the same frequency. hat can be repeated all alng the medium, and s the entire wae thrughut the medium has the same frequency as the surce. 3. he speed f the transerse wae is measuring hw fast the wae disturbance mes alng the crd. Fr a unifrm crd, that speed is cnstant, and depends n the tensin in the crd and the mass density f the crd. he speed f a tiny piece f the crd is measuring hw fast the piece f crd mes perpendicularly t the crd, as the disturbance passes by. hat speed is nt cnstant if a sinusidal wae is traeling n the crd, the speed f each piece f the crd will be gien by the speed relatinship f a simple harmnic scillatr (Equatin -9), which depends n the amplitude f the wae, the frequency f the wae, and the specific time f bseratin. 4. Frm Equatin -9b, the fundamental frequency f scillatin fr a string with bth ends fixed is f. he speed f waes n the string is gien by Equatin -3, F. Cmbining m 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 70

3 Giancli Physics: Principles with Applicatins, 6 th Editin F these tw relatinships gies f. By wrapping the string with wire, the mass f the string m can be greatly increased withut changing the length r the tensin f the string, and thus the string has a lw fundamental frequency. 5. If yu strike the hrizntal rd ertically, yu will create primarily transerse waes. If yu strike the rd parallel t its length, yu will create primarily lngitudinal waes. 6. Frm Equatin -4b, the speed f waes in a gas is gien by B. A decrease in the density due t a temperature increase therefre leads t a higher speed f sund. We expect the speed f sund t increase as temperature increases. 7. (a) Similar t the discussin in sectin -9 fr spherical waes, as a circular wae expands, the circumference f the wae increases. Fr the energy in the wae t be cnsered, as the circumference increases, the intensity has t decrease. he intensity f the wae is prprtinal t the square f the amplitude (b) he water waes will decrease in amplitude due t dissipatin f energy frm iscsity in the water (dissipatie r frictinal energy lss). 8. Assuming the tw waes are in the same medium, then they will bth hae the same speed. Since f, the wae with the smaller waelength will hae twice the frequency f the ther wae. Frm Equatin -8, the intensity f wae is prprtinal t the square f the frequency f the wae. hus the wae with the shrter waelength will transmit 4 times as much energy as the ther wae. 9. he frequency must stay the same because the media is cntinuus the end f ne sectin f crd is physically tied t the ther sectin f crd. If the end f the first sectin f crd is ibrating up and dwn with a gien frequency, then since it is attached t the ther sectin f crd, the ther sectin must ibrate at the same frequency. If the tw pieces f crd did nt me at the same frequency, they wuld nt stay cnnected, and then the waes wuld nt pass frm ne sectin t anther. 0. he string culd be tuched at the lcatin f a nde withut disturbing the mtin, because the ndes d nt me. A string ibrating in three segments has ndes in additin t the nes at the ends. See the diagram. nde nde. he energy f a wae is nt lcalized at ne pint, because the wae is nt lcalized at ne pint, and s t talk abut the energy at a nde being zer is nt really a meaningful statement. Due t the interference f the waes the ttal energy f the medium particles at the ndes pints is zer, but the energy f the medium is nt zer at pints f the medium that are nt ndes. In fact, the anti-nde pints hae mre energy than they wuld hae if nly ne f the tw waes were present.. A majr distinctin between energy transfer by particles and energy transfer by waes is that particles must trael in a straight line frm ne place t anther in rder t transfer energy, but waes can diffract arund bstacles. Fr instance, sund can be heard arund a crner, while yu cannt 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 7

4 Chapter Vibratins and Waes thrw a ball arund a crner. S if a barrier is placed between the surce f the energy and the lcatin where the energy is being receied, and energy is still receied in spite f the barrier, it is a gd indicatin that the energy is being carried by waes. If the placement f the barrier stps the energy transfer, it culd be that the energy transfer is being carried ut by particles. It culd als be that the energy transfer is being carried ut with waes whse waelength is much smaller than the dimensins f the barrier. Slutins t Prblems. he particle wuld trael fur times the amplitude: frm x A t x 0 t x A t x 0 t x A. S the ttal distance 4A 40.8 m 0.7 m.. he spring cnstant is the rati f applied frce t displacement. F 80 N 75 N 05 N k 5.30 N m x 0.85 m 0.65 m 0.0 m 3. he spring cnstant is fund frm the rati f applied frce t displacement. F mg 68 kg9.8m s 5 k.3330 N m 3 x x 50 m he frequency f scillatin is fund frm the ttal mass and the spring cnstant. 5 k.3330 N m.467 Hz.5 Hz f m 568 kg 4. (a) he spring cnstant is fund frm the rati f applied frce t displacement. F mg.7 kg9.80 m s k 735 N m 7.40 N m x x 3.60 m (b) he amplitude is the distance pulled dwn frm equilibrium, s A.5 0 m he frequency f scillatin is fund frm the ttal mass and the spring cnstant. k 735N m f.66 Hz.6 Hz m.7 kg 5. he spring cnstant is the same regardless f what mass is hung frm the spring. f k m k f m cnstant f m f m f f m m 3.0 Hz 0.60 kg 0.38 kg 3.8 Hz 6. he table f data is shwn, alng with the smthed graph. Eery quarter f a perid, the mass mes frm an extreme pint t the time psitin 0 - A /4 0 / A 3/4 0 - A 5/ time / 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 7

5 Giancli Physics: Principles with Applicatins, 6 th Editin equilibrium. he graph resembles a csine wae (actually, the ppsite f a csine wae). 7. he relatinship between frequency, mass, and spring cnstant is (a) f k f m 4 k f. m k Hz.50 kg N m 0.6 N m m k N m (b) f 4.8 Hz m 5.00 kg 8. he spring cnstant is the same regardless f what mass is attached t the spring. k k f mf cnstant m f m f m kg0.60 Hz m kg0.88 Hz m kg 0.68 kg0.60 Hz m 0.59 kg 0.88 Hz 0.60 Hz 9. (a) At equilibrium, the elcity is its imum. k AA fa 3 Hz0.3 m.450 m s.5m s m (b) Frm Equatin (-5), we find the elcity at any psitin. x 0.0 m.45m s.565m s.6m s A 0.3 m (c) E ttal m 0.60 kg.45 m s.80j.8 J (d) Since the bject has a imum displacement at t = 0, the psitin will be described by the csine functin. 0.3 mcs 3.0 Hz 0.3 mcs 6.0 x t x t 0. he relatinship between the elcity and the psitin f a SHO is gien by Equatin (-5). Set that expressin equal t half the imum speed, and sle fr the displacement. x A x A x A x A x 3A 0.866A Since F kx ma fr an bject attached t a spring, the acceleratin is prprtinal t the displacement (althugh in the ppsite directin), as a xk m. hus the acceleratin will hae half its imum alue where the displacement has half its imum alue, at x 0. he spring cnstant can be fund frm the stretch distance crrespnding t the weight suspended n the spring..6 kg9.80 m s F mg k 8.5 N m x x 0.35 m 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 73

6 Chapter Vibratins and Waes After being stretched further and released, the mass will scillate. It takes ne-quarter f a perid fr the mass t me frm the imum displacement t the equilibrium psitin..6 kg m k 0.8 s N m 3. (a) he ttal energy f an bject in SHM is cnstant. When the psitin is at the amplitude, the speed is zer. Use that relatinship t find the amplitude. E m kx ka tt m 3.0 kg A x 0.55 m s 0.00 m m 6.00 m k 80 N m (b) Again use cnseratin f energy. he energy is all kinetic energy when the bject has its imum elcity. E m kx ka m tt k 80 N m A m 3.0 kg m m s 0.58 m s 4. he spring cnstant is fund frm the rati f applied frce t displacement. F 80.0 N k N m x 0.00 m Assuming that there are n dissipatie frces acting n the ball, the elastic ptential energy in the laded psitin will becme kinetic energy f the ball. k N m E E kx m x 0.00 m 9.43m s i f m 0.80 kg 5. (a) he wrk dne t cmpress a spring is stred as ptential energy. W 3.0 J W kx k 46.7 N m 4. 0 N m x 0. m (b) he distance that the spring was cmpressed becmes the amplitude f its mtin. he k imum acceleratin is gien by a A. Sle this fr the mass. m k k N m a A m A 0. m kg 3.3 kg m a 5m s 6. he general frm f the mtin is x Acst 0.45cs 6.40t. (a) he amplitude is A x 0.45 m s (b) he frequency is fund by f 6.40 s f.09 Hz.0 Hz (c) he ttal energy is gien by E m m A 0.60 kg 6.40 s ttal 0.45 m.488j.5 J (d) he ptential energy is gien by 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 74

7 Giancli Physics: Principles with Applicatins, 6 th Editin 0.60 kg 6.40 s 0.30 m.j. J ptential E kx m x he kinetic energy is gien by E E E.488 J. J.377 J.4 J kinetic ttal ptential 7. If the energy f the SHO is half ptential and half kinetic, then the ptential energy is half the ttal energy. he ttal energy is the ptential energy when the displacement has the alue f the amplitude. E E kx ka pt tt x A 0.707A 8. If the frequencies and masses are the same, then the spring cnstants fr the tw ibratins are the same. he ttal energy is gien by the imum ptential energy. ka E A A E ka A A 9. (a) he general equatin fr SHM is Equatin (-8c), y Acs t. Fr the pumpkin, t y 0.8 mcs 0.65 s. (b) he time t return back t the equilibrium psitin is ne-quarter f a perid. t s 0.6 s (c) he imum speed is gien by the angular frequency times the amplitude. A A 0.8 m.7 m s 0.65 s (d) he imum acceleratin is gien by m 7 m s a A A. he imum acceleratin is first attained at the release pint f the pumpkin s 0. Cnsider the first free-bdy diagram fr the blck while it is at equilibrium, s that the net frce is zer. Newtn s nd law fr ertical frces, chsing up as psitie, gies this. F F F mg 0 F F mg y A B A B Nw cnsider the secnd free-bdy diagram, in which the blck is displaced a distance x frm the equilibrium pint. Each upward frce will hae increased by an amunt kx, since x 0. Again write Newtn s nd law fr ertical frces. F F F F mg F kxf kxmg kx F F mg kx y net A B A B A B his is the general frm f a restring frce that prduces SHM, with an effectie spring cnstant f k. hus the frequency f ibratin is as fllws. F A F B mg F A F B mg x 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 75

8 Chapter Vibratins and Waes k f k m effectie m. he equatin f mtin is x 0.38sin 6.50t Asint. (a) he amplitude is A x 0.38 m s (b) he frequency is fund by f 6.50 s f.03 Hz (c) he perid is the reciprcal f the frequency. f.03 Hz s. (d) he ttal energy is gien by E m m A kg 6.50 s ttal 0.38 m 0.95 J 0.9 J. (e) he ptential energy is gien by E kx m x ptential kg 6.50 s 0.090m 0.053J 5. 0 J. he kinetic energy is gien by E E E 0.95J 0.053J J 0.86 J. kinetic ttal ptential (f) x (m) time (sec). (a) Fr A, the amplitude is A.5 m. Fr B, the amplitude is A 3.5 m. A B (b) Fr A, the frequency is cycle eery 4.0 secnds, s f 0.5 Hz. Fr B, the frequency is A cycle eery.0 secnds, s f 0.50 Hz. B (c) Fr C, the perid is 4.0 s. Fr B, the perid is.0 s A B (d) Object A has a displacement f 0 when t 0, s it is a sine functin. x A sin A A f t A x A.5 msin t Object B has a imum displacement when t 0, s it is a csine functin. x A cs f t x 3.5 m cs t B B B B 3. (a) Find the perid and frequency frm the mass and the spring cnstant. m k kg 4 N m s f s.04 Hz (b) he initial speed is the imum speed, and that can be used t find the amplitude. 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 76

9 Giancli Physics: Principles with Applicatins, 6 th Editin A k m A m k.96 m s kg 4 N m 0.3 m (c) he imum acceleratin can be fund frm the mass, spring cnstant, and amplitude a Ak m 0.3 m4 N m kg 37.9m s (d) Because the mass started at the equilibrium psitin f x = 0, the psitin functin will be prprtinal t the sine functin. x 0.3 msin.04 Hz t x 0.3 msin 4.08 t (e) he imum energy is the kinetic energy that the bject has when at the equilibrium psitin. E m kg.96m s 3.3 J 4. We assume that dwnward is the psitie directin f mtin. Fr this mtin, we hae k 305 N m, A0.80 m, m 0.60 kg and k m 305 N m 0.60 kg rad s. (a) Since the mass has a zer displacement and a psitie elcity at t = 0, the equatin is a sine functin. y t 0.80 msin34.3rad st (b) he perid f scillatin is gien by s. he spring will hae 34.5 rad s its imum extensin at times gien by the fllwing. t n s n 0.83 s, n 0,,, 4 he spring will hae its minimum extensin at times gien by the fllwing. 3 t n.380 s n min 0.83 s, n 0,,, 4 5. If the blck is displaced a distance x t the right in the diagram, then spring # will exert a frce F kx, in the ppsite directin t x. ikewise, spring # will exert a frce F kx, in the same directin as F. hus the net frce n the blck is F F F k xk x k k x. he effectie spring cnstant is thus k k k, and the perid is gien by m m k k k. 6. he energy f the scillatr will be cnsered after the cllisin. hus E ka m M A k m M his speed is the speed that the blck and bullet hae immediately after the cllisin. inear mmentum in ne dimensin will hae been cnsered during the cllisin, and s the initial speed f the bullet can be fund. p p m mm befre after 3 mm k 6.50 kg N m A.50 m 597 m s m mm.50 kg 6.50 kg 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 77

10 Chapter Vibratins and Waes 38.0 s 7. he perid f the jumper s mtin is 4.75 s. he spring cnstant can then be fund 8 cycles frm the perid and the jumper s mass. m 4 m kg k 3.73N m 4 N m k 4.75 s 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 78 he stretch f the bungee crd needs t pride a frce equal t the weight f the jumper when he is at the equilibrium pint kg 9.80 m s mg kxmg x 5.60 m k 3.73N m hus the unstretched bungee crd must be 5.0 m 5.60 m 9.4 m 60 s 8. (a) he perid is gien by.7s cycle. 36 cycles (b) he frequency is gien by 36 cycles f 0.60 Hz. 60 s 9. he perid f a pendulum is gien by g. Sle fr the length using a perid f.0 secnds. g g.0 s 9.8m s 0.99 m he perid f a pendulum is gien by g. he length is assumed t be the same fr the pendulum bth n Mars and n Earth. g Mars Mars gearth g g g g Earth Earth 0.80 s.3 s Earth Mars Earth g 0.37 Mars Mars 3. he perid f a pendulum is gien by g m (a) g.8 s 9.8m s (b) If the pendulum is in free fall, there is n tensin in the string supprting the pendulum bb, and s n restring frce t cause scillatins. hus there will be n perid the pendulum will nt scillate and s n perid can be defined. 3. (a) he frequency can be fund frm the length f the pendulum, and the acceleratin due t graity. g 9.80 m s f Hz 0.57 Hz m (b) find the speed at the lwest pint, use the cnseratin f h cs 0 cs

11 Giancli Physics: Principles with Applicatins, 6 th Editin energy relating the lwest pint t the release pint f the pendulum. ake the lwest pint t be the zer leel f graitatinal ptential energy. E E KE PE KE PE tp bttm tp tp bttm bttm mg m bttm 0 cs 0 g cs 9.80 m s 0.760m cs m s bttm (c) he ttal energy can be fund frm the kinetic energy at the bttm f the mtin. E ttal m kg 0.57m s J bttm 33. here are 4 h 60 min h 60s min 86, 400 s in a day. he clck shuld make ne cycle in exactly tw secnds (a tick and a tck ), and s the clck shuld make 43,00 cycles per day. After ne day, the clck in questin is 30 secnds slw, which means that it has made 5 less cycles than required fr precise timekeeping. hus the clck is nly making 43,85 cycles in a day. Accrdingly, the perid f the clck must be decreased by a factr 43,85 43, ,85 43,85 g new ld new g ld 43, 00 43, 00 43,85 43,85 new ld m0.993 m 43, 00 43, 00 hus the pendulum shuld be shrtened by 0.7 mm. 34. Use energy cnseratin t relate the ptential energy at the imum height f the pendulum t the kinetic energy at the lwest pint f the swing. ake the lwest pint t be the zer lcatin fr graitatinal ptential energy. See the diagram. E E KE PE KE PE tp bttm tp tp bttm bttm 0 mgh m gh g cs h cs 0 cs 35. he equatin f mtin fr an bject in SHM that has the imum displacement at t 0 is gien by x Acs f t. Fr a pendulum, x and s x A, where must be measured in radians. hus the equatin fr the pendulum s angular displacement is cs ft cs ft If bth sides f the equatin are multiplied by 80 rad, then the angles can be measured in degrees. hus the angular displacement f the pendulum can be written as belw. Please nte that the argument f the csine functin is still in radians. cs ft 5 cs 5.0 t (a) t (b) t (c) t 0.5 s 5 cs s 5 cs (here the time is exactly 4 perids) 500 s 5 cs (here the time is exactly 50 perids) 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 79

12 Chapter Vibratins and Waes 36. he wae speed is gien by f. he perid is 3.0 secnds, and the waelength is 6.5 m. f 6.5 m 3.0 s.m s 37. he distance between wae crests is the waelength f the wae. f 343m s 6 Hz.3 m 38. find the waelength, use f m s m s AM: 545 m 88 m AM: 90 m t 550 m 3 3 f 5500Hz f 6000Hz FM: 3.4 m.78 m FM:.78 m t 3.4 m m s m s 6 6 f Hz f 080 Hz 39. he elastic and bulk mduli are taken frm able 9- in chapter 9. he densities are taken frm able 0- in chapter N m 3 (a) Fr water: B.40 m s kg m (b) Fr granite: (c) Fr steel: N m E kg m N m E kg m m s m s 40. he speed f a lngitudinal wae in a slid is gien by E. Call the density f the less dense material, and the density f the mre dense material. he less dense material will hae the higher speed, since the speed is inersely prprtinal t the square rt f the density. E.4 E 4. find the time fr a pulse t trael frm ne end f the crd t the ther, the elcity f the pulse n the crd must be knwn. Fr a crd under tensin, we hae. m x F x 8 m t 0.35 s t m F 50 N m 4. (a) he speed f the pulse is gien by x 60 m 77.5m s 78m s t 6 s 0.65 kg 8 m 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 80 F

13 Giancli Physics: Principles with Applicatins, 6 th Editin (b) he tensin is related t the speed f the pulse by cable can be fund frm its lume and density. F. he mass per unit length f the m m m m d m 7.80 kg m.378 kg m V d F m 3 F 77.5m s.378kg m N m 43. he speed f the water wae is gien by B, where B is the bulk mdulus f water, frm able 9-, and is the density f sea water, frm able 0-. he wae traels twice the depth f the cean during the elapsed time. 9 t t B 3.0 s.00 N m 3.0 m 3 3 t.050 kg m 44. (a) Bth waes trael the same distance, s x t t. We let the smaller speed be, and the larger speed be. he slwer wae will take lnger t arrie, and s t is mre than t. t t.0min t 0 s t 0 s t 5.5km s t 0 s 0 s0 s 8.5km s 5.5km s 3 8.5km s 0 s.9 0 km x t (b) his is nt enugh infrmatin t determine the epicenter. All that is knwn is the distance f the epicenter frm the seismic statin. he directin is nt knwn, s the epicenter lies n a 3 circle f radius.9 0 km frm the seismic statin. Readings frm at least tw ther seismic statins are needed t determine the epicenter s psitin. 45. We assume that the earthquake wae is ming the grund ertically, since it is a transerse wae. An bject sitting n the grund will then be ming with SHM, due t the tw frces n it the nrmal frce upwards frm the grund and the weight dwnwards due t graity. If the bject lses cntact with the grund, then the nrmal frce will be zer, and the nly frce n the bject will be its weight. If the nly frce is the weight, then the bject will hae an acceleratin f g dwnwards. hus the limiting cnditin fr beginning t lse cntact with the grund is when the imum acceleratin caused by the wae is greater than g. Any larger dwnward acceleratin and the grund wuld fall quicker than the bject. he imum acceleratin is related t the amplitude and the frequency as fllws. g g 9.8m s a A g A 0.99 m 4 f Hz 46. (a) Assume that the earthquake waes spread ut spherically frm the surce. Under thse cnditins, Eq. (-6b) applies, stating that intensity is inersely prprtinal t the square f the distance frm the surce f the wae. I I 0 km 0 km 0 km 0 km Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 8

14 Chapter Vibratins and Waes (b) he intensity is prprtinal t the square f the amplitude, and s the amplitude is inersely prprtinal t the distance frm the surce f the wae. A A 0 km 0 km km 0 km 47. (a) Assuming spherically symmetric waes, the intensity will be inersely prprtinal t the square f the distance frm the surce. hus Ir will be cnstant. I r I r near near far far r 6 48 km far 9 9 I.0 0 W m W m 4.60 W m near far rnear km I (b) he pwer passing thrugh an area is the intensity times the area. 9 0 PIA W m 5.0 m.3 0 W 48. Frm Equatin (-8), if the speed, medium density, and frequency f the tw waes are the same, then the intensity is prprtinal t the square f the amplitude. I I E E A A A A.4 he mre energetic wae has the larger amplitude. 49. Frm Equatin (-8), if the speed, medium density, and frequency f the tw waes are the same, then the intensity is prprtinal t the square f the amplitude. I I P P A A 3 A A 3.73 he mre energetic wae has the larger amplitude. 50. he bug mes in SHM as the wae passes. he imum KE f a particle in SHM is the ttal energy, which is gien by E ka. Cmpare the tw KE ima. ttal KE ka A.5 cm KE ka A 3.0 cm (a) (b) (c) he energy is all kinetic energy at the mment when the string has n displacement. here is n elastic ptential energy at that mment. Each piece f the string has speed but n displacement. 5. he frequencies f the harmnics f a string that is fixed at bth ends are gien by f nf, and s n the first fur harmnics are f 440 Hz, f 880 Hz, f 30 Hz, f 760 Hz Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 8

15 Giancli Physics: Principles with Applicatins, 6 th Editin 53. he fundamental frequency f the full string is gien by f 94 Hz. If the length is unfingered reduced t /3 f its current alue, and the elcity f waes n the string is nt changed, then the new frequency will be f f fingered unfingered 94 Hz 44 Hz Fur lps is the standing wae pattern fr the 4 th harmnic, with a frequency gien by f 4f 80 Hz. hus f 70 Hz, f 40 Hz, f 0 Hz and f 350 Hz 3 5 are all ther 4 resnant frequencies. 55. Adjacent ndes are separated by a half-waelength, as examinatin f Figure -40 will shw. 9 m s x m nde f f 475 Hz 56. Since f nf, tw successie ertnes differ by the fundamental frequency, as shwn belw. n n n f f f n f nf f 350 Hz 80 Hz 70 Hz 57. he speed f waes n the string is gien by equatin (-3), f a string with bth ends fixed are gien by equatin (-9b), f n F. he resnant frequencies m n ib, where is the length ib f the prtin that is actually ibrating. Cmbining these relatinships allws the frequencies t be calculated. n F 50 N f f Hz n 3 m 0.6 m 3.60 kg 0.90 m ib f f Hz f 3 f 87.3Hz 3 S the three frequencies are 90 Hz, 580 Hz, 870 Hz, t significant figures. 58. Frm Equatin (-9b), f n n, we see that the frequency is prprtinal t the wae speed n the stretched string. Frm equatin (-3), F, we see that the wae speed is prprtinal m t the square rt f the tensin. hus the frequency is prprtinal t the square rt f the tensin. F f f 00 Hz F F F 0.95 F F f f 05 Hz hus the tensin shuld be decreased by 4.8%. 59. he string must ibrate in a standing wae pattern t hae a certain number f lps. he frequency f the standing waes will all be 60 Hz, the same as the ibratr. hat frequency is als expressed 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 83

16 Chapter Vibratins and Waes by Equatin (-9b), f n n. he speed f waes n the string is gien by Equatin (-3), F. he tensin in the string will be the same as the weight f the masses hung frm the end m f the string, F mg. Cmbining these relatinships gies an expressin fr the masses hung frm the end f the string. n n F n mg 4f m n (a) f m n m m n g (b) (c) m s 4.50 m 60 Hz kg m m.89 kg.3 kg m.89 kg m 0.3 kg 4 m.89 kg m 5. 0 kg he tensin in the string is the weight f the hanging mass, F mg. he speed f waes n the F mg string can be fund by, and the frequency is gien as f 60 Hz. he m m waelength f waes created n the string will thus be gien by kg 9.80 m s 4 mg m. f f m 60 Hz kg m he length f the string must be an integer multiple f half f the waelength fr there t be ndes at bth ends and thus frm a standing wae. hus,, 3,, and s n. his gies 0.37 m, 0.75 m,. m,.49 m as the pssible lengths, and s there are 4 standing wae patterns that may be achieed. 6. Frm the descriptin f the water s behair, there is an anti-nde at each end f the tub, and a nde in the middle. hus ne waelength is twice the tube length. f f 0.65 m 0.85 Hz.m s tub 6. he speed in the secnd medium can be fund frm the law f refractin, Equatin (-0). sin sin sin km s 6.3km s sin sin sin he angle f refractin can be fund frm the law f refractin, Equatin (-0). sin.m s sin sin sin sin sin.8m s 64. he angle f refractin can be fund frm the law f refractin, Equatin (-0). he relatie elcities can be fund frm the relatinship gien in the prblem. 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 84

17 Giancli Physics: Principles with Applicatins, 6 th Editin sin sin sin sin sin 5 sin he angle f refractin can be fund frm the law f refractin, Equatin (-0). he relatie elcities can be fund frm Equatin (-4a). sin E SG SG water sin E SG SG water SG 3.6 sin sin sin sin SG he errr f is allwed due t diffractin f the waes. If the waes are incident at the edge f the dish, they can still diffract int the dish if the relatinship is satisfied. rad 0.5 m.7450 m 0 m 80 If the waelength is lnger than that, there will nt be much diffractin, but shadwing instead. 67. he unusual decrease f water crrespnds t a trugh in Figure -4. he crest r peak f the wae is then ne-half waelength distant. he peak is 5 km away, traeling at 750 km/hr. x 5 km 60 min xt t 0 min 750km hr hr 68. Apply the cnseratin f mechanical energy t the car, calling cnditin # t be befre the cllisin and cnditin # t be after the cllisin. Assume that all f the kinetic energy f the car is cnerted t ptential energy stred in the bumper. We knw that x 0 and 0. E E m kx m kx m kx x m 500 kg. m s 0. m k 5500 N m Cnsider the cnseratin f energy fr the persn. Call the unstretched psitin f the fire net the zer lcatin fr bth elastic ptential energy and graitatinal ptential energy. he amunt f stretch f the fire net is gien by x, measured psitiely in the dwnward directin. he ertical displacement fr graitatinal ptential energy is gien by the ariable y, measured psitiely fr the upward directin. Calculate the spring cnstant by cnsering energy between the windw height and the lwest lcatin f the persn. he persn has n kinetic energy at either lcatin. E E mgy mgy kx tp bttm tp bttm bttm y y tp 8 m. m k mg 65 kg 9.8 m s.00 N m x bttm 4 bttm. m (a) If the persn were t lie n the fire net, they wuld stretch the net an amunt such that the upward frce f the net wuld be equal t their weight. 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 85

18 Chapter Vibratins and Waes 65 kg9.8m s mg F kx mg x 3.0 m 4 k.00 N m (b) find the amunt f stretch gien a starting height f 35 m, again use cnseratin f energy. Nte that y x bttm, and there is n kinetic energy at the tp r bttm psitins. mg mg E E mgy mgy kx x x y 0 tp bttm tp bttm tp k k x 65 kg9.8m s 65 kg9.8 m s x 35 m N m.00 N m x x.73 0 x.5 m,.458 m his is a quadratic equatin. he slutin is the psitie rt, since the net must be belw the unstretched psitin. he result is.5 m. 70. Cnsider energy cnseratin fr the mass er the range f mtin frm letting g (the highest pint) t the lwest pint. he mass falls the same distance that the spring is stretched, and has n KE at either endpint. Call the lwest pint the zer f graitatinal ptential energy. he ariable x represents the amunt that the spring is stretched frm the equilibrium psitin. E E tp bttm m mgy kx m mgy kx tp tp tp bttm bttm bttm k g g 0 mgh kh m H H g 9.8ms f. Hz H 0.33 m x = 0 y = H x = H y = 0 7. (a) Frm cnseratin f energy, the initial kinetic energy f the car will all be changed int elastic ptential energy by cmpressing the spring. E E m kx m kx m kx k m s 4 4 m x 5.0 m 950 kg N m.8 0 N m (b) he car will be in cntact with the spring fr half a perid, as it mes frm the equilibrium lcatin t imum displacement and back t equilibrium. m 950 kg 0.7 s k.8390 N m 4 7. he frequency at which the water is being shaken is abut Hz. he slshing cffee is in a standing wae mde, with anti-ndes at each edge f the cup. he cup diameter is thus a half-waelength, r 6 cm. he wae speed can be calculated frm the frequency and the waelength. f 6 cm Hz 6cm s 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 86

19 Giancli Physics: Principles with Applicatins, 6 th Editin 73. Relatie t the fixed needle psitin, the ripples are ming with a linear elcity gien by re min 0.08 m m s min 60 s re his speed is the speed f the ripple waes ming past the needle. he frequency f the waes is 0.373m s f 0 Hz m 74. he equatin f mtin is x 0.650cs 7.40t Acst. (a) he amplitude is A m (b) he frequency is gien by f 7.40 rad s 7.40 rad s f.77 Hz.8 Hz rad (c) he ttal energy is gien by E ka m A ttal.00 kg7.40 rad s m 3.36 J 3. J. (d) he ptential energy is fund by PE kx m x.00 kg7.40 rad s 0.60 m 3.70 J 3.70 J. he kinetic energy is fund by KE E PE 3.36 J 3.70 J 9.4 J. ttal g 75. he frequency f a simple pendulum is gien by f. he pendulum is accelerating ertically which is equialent t increasing (r decreasing) the acceleratin due t graity by the acceleratin f the pendulum. (a) (b) g a.50g g f f. f new g a 0.5g g f f 0.7 f new 76. he frce f the man s weight causes the raft t sink, and that causes the water t put a larger upward frce n the raft. his extra buyant frce is a restring frce, because it is in the ppsite directin f the frce put n the raft by the man. his is analgus t pulling dwn n a mass-spring system that is in equilibrium, by applying an extra frce. hen when the man steps ff, the restring frce pushes upward n the raft, and thus the raft water system acts like a spring, with a spring cnstant fund as fllws. F 75 kg9.8m s 4 k N m x 4.00 m (a) he frequency f ibratin is determined by the spring cnstant and the mass f the raft. 4 k N m f n.455hz.5 Hz m 0 kg (b) As explained in the text, fr a ertical spring the graitatinal ptential energy can be ignred if the displacement is measured frm the scillatr s equilibrium psitin. he ttal energy is thus E 4 ka N m m 4.7 J 5 J ttal. 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 87

20 Chapter Vibratins and Waes 77. (a) he ertnes are gien by f nf, n,3,4 n f f G : f 39 Hz 784 Hz 3 39 Hz 80 Hz 3 A : f 440 Hz 880 Hz Hz 30 Hz 3 (b) If the tw strings hae the same length, they hae the same waelength. he frequency difference is then due t a difference in wae speed caused by different masses fr the strings. F f m G m m f 440 G G G A G A.6 f m m f 39 A A A F G A G m A (c) If the tw strings hae the same mass per unit length and the same tensin, then the wae speed n bth strings is the same. he frequency difference is then due t a difference in waelength. Fr the fundamental, the waelength is twice the length f the string. f f 440 G G A A G A. f f 39 A A G G A G (d) If the tw strings hae the same length, they hae the same waelength. he frequency difference is then due t a difference in wae speed caused by different tensins fr the strings. F G f m F F f 39 G G G G G G f 440 A A A F F F f A A AA A m 78. (a) Since the crd is nt accelerating t the left r right, the tensin in the crd must be the same eerywhere. hus the tensin is the same in the tw parts f the crd. he speed difference will then be due t the different mass densities f the tw parts f the crd. et the symbl represent the mass per unit length f each part f the crd. F H H F H (b) he waelength rati is fund as fllws. f H H H f H he tw frequencies must be the same fr the crd t remain cntinuus at the bundary. If the tw parts f the crd scillate at different frequencies, the crd cannt stay in ne piece, because the tw parts wuld be ut f phase with each ther at arius times. (c) Since, we see that, and s the waelength is greater in the lighter crd. H H 79. (a) he imum speed is gien by 3 f A 64 Hz.80 m 3.0 m s. (b) he imum acceleratin is gien by a Hz.8 0 m m s 3 3 f A. 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 88

21 Giancli Physics: Principles with Applicatins, 6 th Editin 80. Fr the pebble t lse cntact with the bard means that there is n nrmal frce f the bard n the pebble. If there is n nrmal frce n the pebble, then the nly frce n the pebble is the frce f graity, and the acceleratin f the pebble will be g dwnward, the acceleratin due t graity. his is the imum dwnward acceleratin that the pebble can hae. hus if the bard s dwnward acceleratin exceeds g, then the pebble will lse cntact. he imum acceleratin and the amplitude are related by a f A. 4 g 9.8m s a 4 f A g A. 0 m 4 f 4.5 Hz 8. Fr a resnant cnditin, the free end f the string will be an antinde, and the fixed end f the string will be a nde. he minimum distance frm a nde t an antinde is 4. Other wae patterns that fit the bundary cnditins f a nde at ne end and an antinde at the ther end include 3 4, 5 4,. See the diagrams. he general relatinship is n 4, n,,3,. 4 Sling fr the waelength gies, n,,3, n. 0 n = n = 3 0 n = 5-8. he perid f a pendulum is gien by g, and s the length is (a) (b) (c).000 s 9.793m s gaustin Austin s m s gparis Paris Paris Austin 0.99 m m m 0.99 m m.6 mm.00 s.6 m s gmn Mn m g he spring, riginally f length l 0, will be stretched dwnward t a new equilibrium length when the mass is hung n it. he amunt f dwnward stretch l is fund frm setting the spring frce 0 upward n the mass equal t the weight f the mass: kl0 mg l mg k. he length 0 f the pendulum is then l mg k. he perid f the ertical scillatins is gien by 0 m k, while the perid f the pendulum scillatins is gien by g. Nw er pen cmpare the perids f the tw mtins. 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 89

22 Chapter Vibratins and Waes er l mg k g pen 0 l mg k l k 0 0 mk mg k mg pen er lk 0, by a factr f mg 84. Blck m stays n tp f blck M (executing SHM relatie t the grund) withut slipping due t static frictin. he imum static frictinal frce n m is F mg. his frictinal frce causes fr s blck m t accelerate, s ma mg a g s s. hus fr the blcks t stay in cntact withut slipping, the imum acceleratin f blck M is als a g s. But an bject in SHM k has a imum acceleratin gien by a A A. Equate these tw expressins fr the Mttal imum acceleratin. k g m s s a A g A M m 6.5 kg 0.4 m s M k 30 N m ttal 85. he speed f the pulses is fund frm the tensin and mass per unit length f the wire. F 55 N m s m 0.3 kg 0.0 m he ttal distance traeled by the tw pulses will be the length f the wire. he secnd pulse has a shrter time f trael than the first pulse, by 0.0 ms. d d t t t t m m s.000 t m s d t m s s 6.44 m s he tw pulses meet 6.44 m frm the end where the first pulse riginated. 86. Fr the penny t stay n the blck at all times means that there will be a nrmal frce n the penny frm the blck, exerted upward. If dwn is taken t be the psitie directin, then the net frce n the penny is F mg F ma. Sling fr the magnitude f the nrmal frce gies F mg ma net N N. his expressin is always psitie if the acceleratin is upwards (a < 0 ), and s there is n pssibility f the penny lsing cntact while accelerating upwards. But if a dwnward acceleratin were t be larger than g, then the nrmal frce wuld g t zer, since the nrmal frce cannt switch directins F 0 N. hus the limiting cnditin is a g. his is the imum alue fr the acceleratin. dwn k k Fr SHM, we als knw that a A A A. Equate these tw alues fr the M m M acceleratin. k Mg a A g A M k 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 90

23 Giancli Physics: Principles with Applicatins, 6 th Editin 87. he car n the end f the cable prduces tensin in the cable, and stretches the cable accrding t F Equatin (9-4),, where E is Yung s mdulus. Rearrange this equatin t see that E A EA the tensin frce is prprtinal t the amunt f stretch, F, and s the effectie spring EA cnstant is k. he perid f the buncing can be fund frm the spring cnstant and the mass n the end f the cable. m m 00 kg m 0.40 s k EA N m 3.0 m GA 88. Frm Equatin (9-6) and Figure (9-c), the restring frce n the tp f the Jell-O is F, and is in the ppsite directin t the displacement f the tp frm the equilibrium cnditin. hus GA the spring cnstant fr the restring frce is k. If yu were t lk at a layer f Jell-O clser t the base, the displacement wuld be less, but s wuld the restring frce in prprtin, and s we estimate all f the Jell-O as haing the same spring cnstant. he frequency f ibratin can be determined frm the spring cnstant and the mass f the Jell-O. k GA GA G f m V A 50N m 300 kg m m 3.5 Hz 005 Pearsn Educatin, Inc., Upper Saddle Rier, NJ. All rights resered. his material is prtected under all cpyright laws as they 9